Data on one input file being compared.
示例#1
0
        /// <summary>
        /// Find the difference in 2 arrays of integers.
        /// </summary>
        /// <param name="ArrayA">A-version of the numbers (usualy the old one)</param>
        /// <param name="ArrayB">B-version of the numbers (usualy the new one)</param>
        /// <returns>Returns a array of Items that describe the differences.</returns>
        public static Item[] DiffInt(int[] ArrayA, int[] ArrayB)
        {
            // The A-Version of the data (original data) to be compared.
            DiffData DataA = new DiffData(ArrayA);

            // The B-Version of the data (modified data) to be compared.
            DiffData DataB = new DiffData(ArrayB);

            LCS(DataA, 0, DataA.Length, DataB, 0, DataB.Length);
            return CreateDiffs(DataA, DataB);
        }
示例#2
0
        /// <summary>Scan the tables of which lines are inserted and deleted,
        /// producing an edit script in forward order.  
        /// </summary>
        /// dynamic array
        private static Item[] CreateDiffs(DiffData DataA, DiffData DataB)
        {
            ArrayList a = new ArrayList();
            Item aItem;
            Item[] result;

            int StartA, StartB;
            int LineA, LineB;

            LineA = 0;
            LineB = 0;
            while(LineA < DataA.Length || LineB < DataB.Length) {
                if((LineA < DataA.Length) && (!DataA.modified[LineA])
                  && (LineB < DataB.Length) && (!DataB.modified[LineB])) {
                    // equal lines
                    LineA++;
                    LineB++;

                }
                else {
                    // maybe deleted and/or inserted lines
                    StartA = LineA;
                    StartB = LineB;

                    while(LineA < DataA.Length && (LineB >= DataB.Length || DataA.modified[LineA]))
                        // while (LineA < DataA.Length && DataA.modified[LineA])
                        LineA++;

                    while(LineB < DataB.Length && (LineA >= DataA.Length || DataB.modified[LineB]))
                        // while (LineB < DataB.Length && DataB.modified[LineB])
                        LineB++;

                    if((StartA < LineA) || (StartB < LineB)) {
                        // store a new difference-item
                        aItem = new Item();
                        aItem.StartA = StartA;
                        aItem.StartB = StartB;
                        aItem.deletedA = LineA - StartA;
                        aItem.insertedB = LineB - StartB;
                        a.Add(aItem);
                    } // if
                } // if
            } // while

            result = new Item[a.Count];
            a.CopyTo(result);

            return (result);
        }
示例#3
0
        /// <summary>
        /// Find the difference in 2 text documents, comparing by textlines.
        /// The algorithm itself is comparing 2 arrays of numbers so when comparing 2 text documents
        /// each line is converted into a (hash) number. This hash-value is computed by storing all
        /// textlines into a common hashtable so i can find dublicates in there, and generating a 
        /// new number each time a new textline is inserted.
        /// </summary>
        /// <param name="TextA">A-version of the text (usualy the old one)</param>
        /// <param name="TextB">B-version of the text (usualy the new one)</param>
        /// <param name="trimSpace">When set to true, all leading and trailing whitespace characters are stripped out before the comparation is done.</param>
        /// <param name="ignoreSpace">When set to true, all whitespace characters are converted to a single space character before the comparation is done.</param>
        /// <param name="ignoreCase">When set to true, all characters are converted to their lowercase equivivalence before the comparation is done.</param>
        /// <returns>Returns a array of Items that describe the differences.</returns>
        public static Item[] DiffText(string TextA, string TextB, bool trimSpace, bool ignoreSpace, bool ignoreCase)
        {
            // prepare the input-text and convert to comparable numbers.
            Hashtable h = new Hashtable(TextA.Length + TextB.Length);

            // The A-Version of the data (original data) to be compared.
            DiffData DataA = new DiffData(DiffCodes(TextA, h, trimSpace, ignoreSpace, ignoreCase));

            // The B-Version of the data (modified data) to be compared.
            DiffData DataB = new DiffData(DiffCodes(TextB, h, trimSpace, ignoreSpace, ignoreCase));

            h = null; // free up hashtable memory (maybe)

            LCS(DataA, 0, DataA.Length, DataB, 0, DataB.Length);
            return CreateDiffs(DataA, DataB);
        }
示例#4
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        }         // SMS

        /// <summary>
        /// This is the divide-and-conquer implementation of the longes common-subsequence (LCS)
        /// algorithm.
        /// The published algorithm passes recursively parts of the A and B sequences.
        /// To avoid copying these arrays the lower and upper bounds are passed while the sequences stay constant.
        /// </summary>
        /// <param name="DataA">sequence A</param>
        /// <param name="LowerA">lower bound of the actual range in DataA</param>
        /// <param name="UpperA">upper bound of the actual range in DataA (exclusive)</param>
        /// <param name="DataB">sequence B</param>
        /// <param name="LowerB">lower bound of the actual range in DataB</param>
        /// <param name="UpperB">upper bound of the actual range in DataB (exclusive)</param>
        private static void LCS(DiffData DataA, int LowerA, int UpperA, DiffData DataB, int LowerB, int UpperB)
        {
            // Debug.Write(2, "LCS", String.Format("Analyse the box: A[{0}-{1}] to B[{2}-{3}]", LowerA, UpperA, LowerB, UpperB));

            // Fast walkthrough equal lines at the start
            while (LowerA < UpperA && LowerB < UpperB && DataA.data[LowerA] == DataB.data[LowerB])
            {
                LowerA++; LowerB++;
            }

            // Fast walkthrough equal lines at the end
            while (LowerA < UpperA && LowerB < UpperB && DataA.data[UpperA - 1] == DataB.data[UpperB - 1])
            {
                --UpperA; --UpperB;
            }

            if (LowerA == UpperA)
            {
                // mark as inserted lines.
                while (LowerB < UpperB)
                {
                    DataB.modified[LowerB++] = true;
                }
            }
            else if (LowerB == UpperB)
            {
                // mark as deleted lines.
                while (LowerA < UpperA)
                {
                    DataA.modified[LowerA++] = true;
                }
            }
            else
            {
                // Find the middle snakea and length of an optimal path for A and B
                SMSRD smsrd = SMS(DataA, LowerA, UpperA, DataB, LowerB, UpperB);
                // Debug.Write(2, "MiddleSnakeData", String.Format("{0},{1}", smsrd.x, smsrd.y));

                // The path is from LowerX to (x,y) and (x,y) ot UpperX
                LCS(DataA, LowerA, smsrd.x, DataB, LowerB, smsrd.y);
                LCS(DataA, smsrd.x, UpperA, DataB, smsrd.y, UpperB); // 2002.09.20: no need for 2 points
            }
        }                                                            // LCS()
示例#5
0
        }         // DiffCodes

        /// <summary>
        /// This is the algorithm to find the Shortest Middle Snake (SMS).
        /// </summary>
        /// <param name="DataA">sequence A</param>
        /// <param name="LowerA">lower bound of the actual range in DataA</param>
        /// <param name="UpperA">upper bound of the actual range in DataA (exclusive)</param>
        /// <param name="DataB">sequence B</param>
        /// <param name="LowerB">lower bound of the actual range in DataB</param>
        /// <param name="UpperB">upper bound of the actual range in DataB (exclusive)</param>
        /// <returns>a MiddleSnakeData record containing x,y and u,v</returns>
        private static SMSRD SMS(DiffData DataA, int LowerA, int UpperA, DiffData DataB, int LowerB, int UpperB)
        {
            SMSRD ret;
            int   MAX = DataA.Length + DataB.Length + 1;

            int DownK = LowerA - LowerB;           // the k-line to start the forward search
            int UpK   = UpperA - UpperB;           // the k-line to start the reverse search

            int  Delta    = (UpperA - LowerA) - (UpperB - LowerB);
            bool oddDelta = (Delta & 1) != 0;

            // vector for the (0,0) to (x,y) search
            int[] DownVector = new int[2 * MAX + 2];

            // vector for the (u,v) to (N,M) search
            int[] UpVector = new int[2 * MAX + 2];

            // The vectors in the publication accepts negative indexes. the vectors implemented here are 0-based
            // and are access using a specific offset: UpOffset UpVector and DownOffset for DownVektor
            int DownOffset = MAX - DownK;
            int UpOffset   = MAX - UpK;

            int MaxD = ((UpperA - LowerA + UpperB - LowerB) / 2) + 1;

            // Debug.Write(2, "SMS", String.Format("Search the box: A[{0}-{1}] to B[{2}-{3}]", LowerA, UpperA, LowerB, UpperB));

            // init vectors
            DownVector[DownOffset + DownK + 1] = LowerA;
            UpVector[UpOffset + UpK - 1]       = UpperA;

            for (int D = 0; D <= MaxD; D++)
            {
                // Extend the forward path.
                for (int k = DownK - D; k <= DownK + D; k += 2)
                {
                    // Debug.Write(0, "SMS", "extend forward path " + k.ToString());

                    // find the only or better starting point
                    int x, y;
                    if (k == DownK - D)
                    {
                        x = DownVector[DownOffset + k + 1];                         // down
                    }
                    else
                    {
                        x = DownVector[DownOffset + k - 1] + 1;                         // a step to the right
                        if ((k < DownK + D) && (DownVector[DownOffset + k + 1] >= x))
                        {
                            x = DownVector[DownOffset + k + 1];                             // down
                        }
                    }
                    y = x - k;

                    // find the end of the furthest reaching forward D-path in diagonal k.
                    while ((x < UpperA) && (y < UpperB) && (DataA.data[x] == DataB.data[y]))
                    {
                        x++; y++;
                    }
                    DownVector[DownOffset + k] = x;

                    // overlap ?
                    if (oddDelta && (UpK - D < k) && (k < UpK + D))
                    {
                        if (UpVector[UpOffset + k] <= DownVector[DownOffset + k])
                        {
                            ret.x = DownVector[DownOffset + k];
                            ret.y = DownVector[DownOffset + k] - k;
                            // ret.u = UpVector[UpOffset + k];      // 2002.09.20: no need for 2 points
                            // ret.v = UpVector[UpOffset + k] - k;
                            return(ret);
                        }         // if
                    }             // if
                }                 // for k

                // Extend the reverse path.
                for (int k = UpK - D; k <= UpK + D; k += 2)
                {
                    // Debug.Write(0, "SMS", "extend reverse path " + k.ToString());

                    // find the only or better starting point
                    int x, y;
                    if (k == UpK + D)
                    {
                        x = UpVector[UpOffset + k - 1];                         // up
                    }
                    else
                    {
                        x = UpVector[UpOffset + k + 1] - 1;                         // left
                        if ((k > UpK - D) && (UpVector[UpOffset + k - 1] < x))
                        {
                            x = UpVector[UpOffset + k - 1]; // up
                        }
                    }                                       // if
                    y = x - k;

                    while ((x > LowerA) && (y > LowerB) && (DataA.data[x - 1] == DataB.data[y - 1]))
                    {
                        x--; y--;                         // diagonal
                    }
                    UpVector[UpOffset + k] = x;

                    // overlap ?
                    if (!oddDelta && (DownK - D <= k) && (k <= DownK + D))
                    {
                        if (UpVector[UpOffset + k] <= DownVector[DownOffset + k])
                        {
                            ret.x = DownVector[DownOffset + k];
                            ret.y = DownVector[DownOffset + k] - k;
                            // ret.u = UpVector[UpOffset + k];     // 2002.09.20: no need for 2 points
                            // ret.v = UpVector[UpOffset + k] - k;
                            return(ret);
                        } // if
                    }     // if
                }         // for k
            }             // for D

            throw new ApplicationException("the algorithm should never come here.");
        }         // SMS
示例#6
0
        /// <summary>
        /// This is the algorithm to find the Shortest Middle Snake (SMS).
        /// </summary>
        /// <param name="DataA">sequence A</param>
        /// <param name="LowerA">lower bound of the actual range in DataA</param>
        /// <param name="UpperA">upper bound of the actual range in DataA (exclusive)</param>
        /// <param name="DataB">sequence B</param>
        /// <param name="LowerB">lower bound of the actual range in DataB</param>
        /// <param name="UpperB">upper bound of the actual range in DataB (exclusive)</param>
        /// <returns>a MiddleSnakeData record containing x,y and u,v</returns>
        private static SMSRD SMS(DiffData DataA, int LowerA, int UpperA, DiffData DataB, int LowerB, int UpperB)
        {
            SMSRD ret;
            int MAX = DataA.Length + DataB.Length + 1;

            int DownK = LowerA - LowerB; // the k-line to start the forward search
            int UpK = UpperA - UpperB; // the k-line to start the reverse search

            int Delta = (UpperA - LowerA) - (UpperB - LowerB);
            bool oddDelta = (Delta & 1) != 0;

            // vector for the (0,0) to (x,y) search
            int[] DownVector = new int[2 * MAX + 2];

            // vector for the (u,v) to (N,M) search
            int[] UpVector = new int[2 * MAX + 2];

            // The vectors in the publication accepts negative indexes. the vectors implemented here are 0-based
            // and are access using a specific offset: UpOffset UpVector and DownOffset for DownVektor
            int DownOffset = MAX - DownK;
            int UpOffset = MAX - UpK;

            int MaxD = ((UpperA - LowerA + UpperB - LowerB) / 2) + 1;

            // Debug.Write(2, "SMS", String.Format("Search the box: A[{0}-{1}] to B[{2}-{3}]", LowerA, UpperA, LowerB, UpperB));

            // init vectors
            DownVector[DownOffset + DownK + 1] = LowerA;
            UpVector[UpOffset + UpK - 1] = UpperA;

            for(int D = 0; D <= MaxD; D++) {

                // Extend the forward path.
                for(int k = DownK - D; k <= DownK + D; k += 2) {
                    // Debug.Write(0, "SMS", "extend forward path " + k.ToString());

                    // find the only or better starting point
                    int x, y;
                    if(k == DownK - D) {
                        x = DownVector[DownOffset + k + 1]; // down
                    }
                    else {
                        x = DownVector[DownOffset + k - 1] + 1; // a step to the right
                        if((k < DownK + D) && (DownVector[DownOffset + k + 1] >= x))
                            x = DownVector[DownOffset + k + 1]; // down
                    }
                    y = x - k;

                    // find the end of the furthest reaching forward D-path in diagonal k.
                    while((x < UpperA) && (y < UpperB) && (DataA.data[x] == DataB.data[y])) {
                        x++; y++;
                    }
                    DownVector[DownOffset + k] = x;

                    // overlap ?
                    if(oddDelta && (UpK - D < k) && (k < UpK + D)) {
                        if(UpVector[UpOffset + k] <= DownVector[DownOffset + k]) {
                            ret.x = DownVector[DownOffset + k];
                            ret.y = DownVector[DownOffset + k] - k;
                            // ret.u = UpVector[UpOffset + k];      // 2002.09.20: no need for 2 points
                            // ret.v = UpVector[UpOffset + k] - k;
                            return (ret);
                        } // if
                    } // if

                } // for k

                // Extend the reverse path.
                for(int k = UpK - D; k <= UpK + D; k += 2) {
                    // Debug.Write(0, "SMS", "extend reverse path " + k.ToString());

                    // find the only or better starting point
                    int x, y;
                    if(k == UpK + D) {
                        x = UpVector[UpOffset + k - 1]; // up
                    }
                    else {
                        x = UpVector[UpOffset + k + 1] - 1; // left
                        if((k > UpK - D) && (UpVector[UpOffset + k - 1] < x))
                            x = UpVector[UpOffset + k - 1]; // up
                    } // if
                    y = x - k;

                    while((x > LowerA) && (y > LowerB) && (DataA.data[x - 1] == DataB.data[y - 1])) {
                        x--; y--; // diagonal
                    }
                    UpVector[UpOffset + k] = x;

                    // overlap ?
                    if(!oddDelta && (DownK - D <= k) && (k <= DownK + D)) {
                        if(UpVector[UpOffset + k] <= DownVector[DownOffset + k]) {
                            ret.x = DownVector[DownOffset + k];
                            ret.y = DownVector[DownOffset + k] - k;
                            // ret.u = UpVector[UpOffset + k];     // 2002.09.20: no need for 2 points
                            // ret.v = UpVector[UpOffset + k] - k;
                            return (ret);
                        } // if
                    } // if

                } // for k

            } // for D

            throw new ApplicationException("the algorithm should never come here.");
        }
示例#7
0
        /// <summary>
        /// This is the divide-and-conquer implementation of the longes common-subsequence (LCS) 
        /// algorithm.
        /// The published algorithm passes recursively parts of the A and B sequences.
        /// To avoid copying these arrays the lower and upper bounds are passed while the sequences stay constant.
        /// </summary>
        /// <param name="DataA">sequence A</param>
        /// <param name="LowerA">lower bound of the actual range in DataA</param>
        /// <param name="UpperA">upper bound of the actual range in DataA (exclusive)</param>
        /// <param name="DataB">sequence B</param>
        /// <param name="LowerB">lower bound of the actual range in DataB</param>
        /// <param name="UpperB">upper bound of the actual range in DataB (exclusive)</param>
        private static void LCS(DiffData DataA, int LowerA, int UpperA, DiffData DataB, int LowerB, int UpperB)
        {
            // Debug.Write(2, "LCS", String.Format("Analyse the box: A[{0}-{1}] to B[{2}-{3}]", LowerA, UpperA, LowerB, UpperB));

            // Fast walkthrough equal lines at the start
            while(LowerA < UpperA && LowerB < UpperB && DataA.data[LowerA] == DataB.data[LowerB]) {
                LowerA++; LowerB++;
            }

            // Fast walkthrough equal lines at the end
            while(LowerA < UpperA && LowerB < UpperB && DataA.data[UpperA - 1] == DataB.data[UpperB - 1]) {
                --UpperA; --UpperB;
            }

            if(LowerA == UpperA) {
                // mark as inserted lines.
                while(LowerB < UpperB)
                    DataB.modified[LowerB++] = true;

            }
            else if(LowerB == UpperB) {
                // mark as deleted lines.
                while(LowerA < UpperA)
                    DataA.modified[LowerA++] = true;

            }
            else {
                // Find the middle snakea and length of an optimal path for A and B
                SMSRD smsrd = SMS(DataA, LowerA, UpperA, DataB, LowerB, UpperB);
                // Debug.Write(2, "MiddleSnakeData", String.Format("{0},{1}", smsrd.x, smsrd.y));

                // The path is from LowerX to (x,y) and (x,y) ot UpperX
                LCS(DataA, LowerA, smsrd.x, DataB, LowerB, smsrd.y);
                LCS(DataA, smsrd.x, UpperA, DataB, smsrd.y, UpperB);  // 2002.09.20: no need for 2 points
            }
        }